# Surface integrals

## Homework Statement

Evaluate integral A.n dS for A=(y,2x,-z) and S is the surface of the plane 2x+y=6 in the first octant of the plane cut off by z=4

Integral A.n dS

## The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).
A.n= 1/sqrt3(2y,2x,0)

So this is where I get stuck. I think it goes as follows....

ds= dxdy

=> double integral 2/sqrt3 y + 2/sqrt3 x dxdy

And I'm not sure how to find the limits....

HallsofIvy
Homework Helper
The plane is in the first octant so bounded by x= y= z= 0. Since there is no "z" in the equation of the plane, the plane is parallel to the z- axis so we need another z bound: we are told that the plane is bounded by z= 4. The line 2x+ y= 6 crosses the x-axis at (0, 6) and the y axis at (3, 0). We can take x going from 0 to 3 and, for each x, y from 0 to 6- 2x. Equivalently, we can take y going from 0 to 6 and, for each y, x= (6- y)/2. In either case, z goes from 0 to 4.

The plane is in the first octant so bounded by x= y= z= 0. Since there is no "z" in the equation of the plane, the plane is parallel to the z- axis so we need another z bound: we are told that the plane is bounded by z= 4. The line 2x+ y= 6 crosses the x-axis at (0, 6) and the y axis at (3, 0). We can take x going from 0 to 3 and, for each x, y from 0 to 6- 2x. Equivalently, we can take y going from 0 to 6 and, for each y, x= (6- y)/2. In either case, z goes from 0 to 4.
Would there not only be two bounds as its a surface integral? Forgive me if I'm being thick....

HallsofIvy
Homework Helper
Yes, the bounds on x and y. I didn't say you had to use the z bounds!

LCKurtz
Homework Helper
Gold Member
I would add that the problem is incompletely stated without the orientation of the surface being given.

LCKurtz
Homework Helper
Gold Member
Upon looking more closely at this problem and proposed solution -- you cannot paramaterize this surface in terms of the variables ##x## and ##y## because they are dependent on each other. I would suggest the parameterization ##x = x,~y=6-2x,~z = z##. So the surface is parameterized as ##\vec R(x,z) = \langle x, 6-2x, z\rangle## and use the formula$$\iint_S \vec A\cdot d\vec S = \pm\int\int \vec A \cdot \vec R_x\times \vec R_z~dxdz$$where the limits are over the ##xz## rectangle and the sign is chosen for the correct orientation (which needs to be specified).

pardon me. i faced a similar prblm . could any1 pls xplain what is first octant??

HallsofIvy
Homework Helper
The "first octant" is the octant in an xyz- coordinate system in which all coordinates are positive.

## Homework Statement

Evaluate integral A.n dS for A=(y,2x,-z) and S is the surface of the plane 2x+y=6 in the first octant of the plane cut off by z=4
The "first octant of the plane" doesn't make sense. A plane has four quadrants not octants. I presume that you mean the portion in the first octant of the xyz coordinate, x> 0, y> 0, 2x+ y< 6 and 0< z< 4.

Integral A.n dS

## The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).[/quote]
No, the unit normal is 1/sqrt(5)(2, 1, 0).

A.n= 1/sqrt3(2y,2x,0)

So this is where I get stuck. I think it goes as follows....

ds= dxdy

=> double integral 2/sqrt3 y + 2/sqrt3 x dxdy
Since dS includes the factor sqrt(5) (not sqrt(3)) itself, there is really no point in using "n" at all. I would write, rather, $\vec{A}\cdot \vec{dS}$ where $\vec{dS}= (2, 1, 0)dxdy$ so that the integral will be $\int_{x= 0}^3 \int_{y= 0}^{6- 2x} 2y+ 2x dydx$

And I'm not sure how to find the limits....
In the x,y plane, the line 2x+ y= 6 crosses the x-axis at (3, 0) and crosses the y-axis at (0, 6). The gives a triangle with vertices at (0, 0), (3, 0), and (0, 6) in the first quadrant of the xy- plane. You can take x going from 0 to 3 and, for each x, y from 0 to 6- 2x.

pardon me. i faced a similar prblm . could any1 pls xplain what is first octant??
thnx

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Evaluate integral A.n dS for A=(y,2x,-z) and S is the surface of the plane 2x+y=6 in the first octant of the plane cut off by z=4

Integral A.n dS

## The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).
A.n= 1/sqrt3(2y,2x,0)

So this is where I get stuck. I think it goes as follows....

ds= dxdy

=> double integral 2/sqrt3 y + 2/sqrt3 x dxdy

And I'm not sure how to find the limits....
In your expression ##\vec{A} \cdot \vec{n} \, dS##, ##dS## is the surface area-element in the plane ##2x+y=6##. Thus ##dS = ds \, dz,## where is ##ds = \sqrt{(dx)^2 + (dy)^2}## is the length-element along the surface perpendicular to the ##z##-axis.

LCKurtz
Homework Helper
Gold Member
The "first octant" is the octant in an xyz- coordinate system in which all coordinates are positive.

The "first octant of the plane" doesn't make sense. A plane has four quadrants not octants. I presume that you mean the portion in the first octant of the xyz coordinate, x> 0, y> 0, 2x+ y< 6 and 0< z< 4.

Integral A.n dS

## The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).
No, the unit normal is 1/sqrt(5)(2, 1, 0).

Since dS includes the factor sqrt(5) (not sqrt(3)) itself, there is really no point in using "n" at all. I would write, rather, $\vec{A}\cdot \vec{dS}$ where $\vec{dS}= (2, 1, 0)dxdy$ so that the integral will be $\int_{x= 0}^3 \int_{y= 0}^{6- 2x} 2y+ 2x dydx$
This is not correct. As I explained in post #6, this cannot be worked as an xy domain problem.